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\hskip 1.5in \textbf{Midterm 3A--Stat 100--Spring 2001}   
\vskip .2in  
You will need a calculator. 
You may use ONE page of notes (writing/xeroxing allowed  
both sides). Each problem is 
worth 20 points. Different parts of a problem have equal weight unless 
otherwise indicated.  
 Put a box around the final answer to a question. 
\vskip .2in 

\renewcommand{\labelenumi}{\alph{enumi}.}

***NOTE: When you are asked for a confidence interval, 
do NOT   take 
the time to evaluate numbers on your calculator. 
Just write  out the appropriate formula, 
with all   appropriate numbers substituted.***

\vspace{.3in} 

1.
A university official wishes to estimate the proportion $p$ of students 
who hold part-time jobs.  In each of the two cases 
below, determine how many students should be surveyed if the
official wants to be 95\% certain that the estimate does not differ from
the true proportion by more than .05 .  

\vspace{.1in}

(a) The official is confident  that $p$ is at least .85 . 

(b) The official has no idea about the value of $p$. 
\vspace{.5in}

2. 
The following summary statistics are recorded for  independent samples
from two populations:



\begin{longtable}
{|p{110pt}|p{110pt}|p{110pt}|} a & a & a   \kill 
\hline
	 & Population 1 & Population 2 \\ 
\hline Sample Size & 12    & 14 
\hline Mean & 41.2  & 37.4 \\
\hline Standard Deviation & 4.98 & 3.12  \\
\hline
\end{longtable}
\\

(a) (14 points) Construct a 95\% confidence interval for $\mu_1-\mu_2$. 


(b) (4 points) State any assumptions needed to justify your procedure.

(c) (2 points) 
Is your confidence interval a conservative confidence interval? 

\vspace{.5in}





3. A group of 88 subjects is used in an experiment to compare two
treatments. From this group, 40 subjects are randomly selected to
be assigned to treatment 1 and the remaining 48 subjects are
assigned to treatment 2. The means and standard deviations of the
responses to the treatments are

\vspace{.1in}

\begin{longtable}
{|p{110pt}|p{110pt}|p{110pt}|} a & a & a  \kill 
\hline
           & Treatment 1 & Treatment 2 \\
\hline Mean& 16.21& 27.84 \\
\hline Standard Deviation & 2.88 & 4.32 \\
\hline
\end{longtable}
Determine a 95\% confidence interval for $\mu_1 - \mu_2$, 
where $\mu_1$ is the mean response  to Treatment 1 and 
$\mu_2$ is the mean response to Treatment 2. 





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**************THERE ARE MORE PROBLEMS ON THE BACK SIDE***************
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\hspace{6in} .  
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4. Five years ago, 
a survey showed that the average height of an NBA player was 
78 inches. In a random sample of 25 players of today, the mean height and 
standard deviation is found to be 79.5 and 3 inches, respectively. Is this 
strong evidence that today's mean height 
$\mu$ is greater than the mean height of five years ago? 
Test at level of signifcance $\alpha = .05$ . 
  In your response,

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\begin{enumerate}
\item
Determine the null and alternative hypotheses.
\item
Define the appropriate test statistic. 
\item
Determine the rejection region.
\item
Determine the value of the test statistic.
\item
State your conclusion and estimate a P-value.
\end{enumerate}

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5. Answer each part TRUE or FALSE. 

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 (a)
 If a hypothesis test is run at $\alpha = .05$ and the
P-value is $.03$, then the decision must be to retain 
the null hypothesis. 

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 (b) For a hypothesis test: 
if the probability of a type I error is $\alpha = .05$ , then 
the probability of a type II error is $.95$. 

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 (c) If a hypothesis test is run  at significance level $\alpha = .05$ 
on many samples from a population for which the null hypothesis 
is true, 
 then for about $95\%$ of those samples 
 the conclusion will be to reject the null hypothesis. 

\vspace{.1in}

 (d) 
The point of assuming equal population standard deviations, 
when computing  
a confidence interval for $\mu_1-\mu_2$ 
using the pooled standard deviation estimator $s_p$,
is to produce a smaller 
confidence interval. 




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